Problem: Solve for $x$ and $y$ using elimination. ${-6x+2y = 14}$ ${3x+3y = 33}$
Answer: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Multiply the bottom equation by $2$ ${-6x+2y = 14}$ $6x+6y = 66$ Add the top and bottom equations together. $8y = 80$ $\dfrac{8y}{{8}} = \dfrac{80}{{8}}$ ${y = 10}$ Now that you know ${y = 10}$ , plug it back into $\thinspace {-6x+2y = 14}\thinspace$ to find $x$ ${-6x + 2}{(10)}{= 14}$ $-6x+20 = 14$ $-6x+20{-20} = 14{-20}$ $-6x = -6$ $\dfrac{-6x}{{-6}} = \dfrac{-6}{{-6}}$ ${x = 1}$ You can also plug ${y = 10}$ into $\thinspace {3x+3y = 33}\thinspace$ and get the same answer for $x$ : ${3x + 3}{(10)}{= 33}$ ${x = 1}$